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Editorial Team
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Asked: 2026-06-07T10:33:15+00:00

inline void insert(node *root, int value) { if(!root) { root = new node(); root->value

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inline void insert(node *root, int value)
{
    if(!root)
    {
        root = new node();
        root->value = value;
    }
    else
    {
        node *itr = root;
        while(1)
        {
            if(itr->value > value)
                itr = itr->left;
            else
                itr = itr->right;
            if(!itr)
            {
                itr = new node();
                itr->value = value; 

                break;
            }

        }
    }
}

//call insert function like this

node *head = 0;
insert(head, 5);
insert(head, 10);
insert(head, 3);
insert(head, 1);
insert(head, 4);

I know this code wouldn’t work because ‘itr’ in insert function is a local variable, thus, it wouldn’t reflect the tree outside of the method.
However, I don’t clearly understand why it wouldn’t work.
Although ‘itr’ is a local variable, ‘itr’ points to the same location where ‘root’ points to. Also, I am dereferencing it to move ‘left’ or ‘right’ so I think it should work.

I think this is a basic problem of passing a pointer by value vs pointer but I couldn’t find a clear explanation of why I can’t change the tree using a local variable of a pointer.

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1 Answer

  1. Editorial Team

    Suppose you had

    int x = 0;
int y = x;
y = 3478;

    Would you expect that x also contain 3478?
    I knew you wouldn’t, and the same is true for your root and itr.

    This is a classic pencil-and-paper problem (most pointer problems are) and it’s definitely worth pulling out some dead tree when you run into problems like this.

    Here’s an ASCII version for one of your cases, where you want to insert to the right, and right is NULL.
    The arrows show where the respective variables are pointing.

    Start of function:

               ____
root ---> |    |
          ------
           / \
          /   \
       left   NULL

    itr = root;    
           ____
root ---> |    | <--- itr
          ------
           / \
          /   \
       left   NULL

    itr = itr->right;
           ____
root ---> |    | 
          ------
           / \
          /   \
       left   NULL <--- itr

    if (!itr)
    itr = new node();

           ____
root ---> |    | 
          ------
           / \
          /   \                   ____
       left   NULL      itr ---> |    |
                                  ----

    As you can see, the input tree hasn’t been modified at all, you just allocated a new node outside of it and left it there.

    This would work:

               ____
root ---> |    | <--- itr
          ------
           / \
          /   \
       left   NULL

       if (!itr->right)
   {
       itr->right = new node()
   }


           ____
root ---> |    | <--- itr
          ------
           / \
          /   \
       left   ____
             |    |
              ----

    Pencil and paper is the best way to figure out pointers.

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